w pewnym roztworze wodorotlenku baru stezenie jonow Ba^2+ wynosi 5* 10^-4mol/ dm3. oblicz pH tego roztworu.
[OH] = 2* [Ba] = 2*5*10^-4=0,001 mol/dm^3pOH= -log[OH]= 3pH= 14- pOH = 14-3 = 11Odp. pH= 11
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[OH] = 2* [Ba] = 2*5*10^-4=0,001 mol/dm^3
pOH= -log[OH]= 3
pH= 14- pOH = 14-3 = 11
Odp. pH= 11